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eimsori [14]
3 years ago
11

How does the length and width of a flute affect its sound?

Physics
1 answer:
Butoxors [25]3 years ago
4 0

https://www.yamaha.com/en/musical_instrument_guide/flute/mechanism/mechanism002.html

Your welcome!

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What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so i
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Answer:

Explanation:

Acceleration of gravity on the surface of

Neptune is given

g' = G M/R^2,

where M is the planet's mass.

M g' will be the apparent weight registed at the poles.

G is the universal constant of gravity

The "apparent weight" at the equator will be

reduced from M g' by the centripetal force M

R w^2, where w is the rotational angular

velocity in radians/s), which you can deduce

from the 16 hour length of a day.

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3 years ago
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Which products are formed when ethane (C2H6) undergoes combustion?
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Answer:

C.

Explanation:

It is C.

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3 years ago
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Eva is in a closed, dark room. She uses her arm muscles to turn on a lamp. When she moves her hand closer to the lamp, the light
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Well, we know that the total energy in a closed system remains constant.

The problem with the story of Eva is that she is not in a closed system. 
If the dark room were really a closed system, then she could press the
button or turn the switch all day, and the lamp could not light.  It needs
electrical energy coming in from somewhere in order to turn on.

Let's say that Eva used her arm muscles to strike a match and light the
candle on the table.  Then we would have have food energy, muscle
energy, chemical energy in the match, chemical energy in the candle,
heat and light energy coming out of the candle, heat energy soaking into
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4 0
4 years ago
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A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

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3 years ago
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